Let `f (x) = (cos ^(2) x)/(1+ cos x+cos ^(2)x )and g (x) =lamda tan x+(1-lamda) sin x-x, ` where `lamda in R and x in [0, pi//2).`
The values of `'lamda'` such that `g '(x) ge 0` for any `x in [0, pi//2):`

Options

A. (a) `[1,oo)`
B. (b) `[0,oo)`
C. (c) `[(1)/(2), oo)`
D. (d) `[(1)/(3), oo)`

Answer

To solve the problem, we need to analyze the function \( g(x) \) and find the values of \( \lambda \) such that \( g'(x) \geq 0 \) for all \( x \) in the interval \( [0, \frac{\pi}{2}) \).

### Step 1: Differentiate \( g(x) \)

Given:
\[ g(x) = \lambda \tan x + (1 - \lambda) \sin x - x \]

To find \( g'(x) \), we differentiate \( g(x) \) with respect to \( x \):
\[ g'(x) = \lambda \sec^2 x + (1 - \lambda) \cos x - 1 \]

### Step 2: Rewrite \( g'(x) \)

We can express \( g'(x) \) as:
\[ g'(x) = \frac{\lambda}{\cos^2 x} + (1 - \lambda) \cos x - 1 \]

### Step 3: Find a common denominator

To combine the terms, we will use a common denominator:
\[ g'(x) = \frac{\lambda + (1 - \lambda) \cos^3 x - \cos^2 x}{\cos^2 x} \]

### Step 4: Factor \( g'(x) \)

Rearranging gives:
\[ g'(x) = \frac{(1 - \lambda) \cos^3 x + \lambda - \cos^2 x}{\cos^2 x} \]

### Step 5: Set the condition for \( g'(x) \geq 0 \)

For \( g'(x) \geq 0 \), the numerator must be non-negative:
\[ (1 - \lambda) \cos^3 x + \lambda - \cos^2 x \geq 0 \]

### Step 6: Analyze the behavior of \( g'(x) \)

1. At \( x = 0 \):
\[ g'(0) = (1 - \lambda) \cdot 1^3 + \lambda - 1 = 1 - \lambda + \lambda - 1 = 0 \]

2. At \( x = \frac{\pi}{2} \):
\[ g'(\frac{\pi}{2}) \text{ is not defined since } \tan x \text{ and } \sec^2 x \text{ are undefined.} \]

### Step 7: Check the critical points in \( (0, \frac{\pi}{2}) \)

We need to check the sign of \( g'(x) \) in the interval. The expression must be non-negative for all \( x \) in \( [0, \frac{\pi}{2}) \).

### Step 8: Determine the bounds of \( f(x) \)

Given:
\[ f(x) = \frac{\cos^2 x}{1 + \cos x + \cos^2 x} \]

1. At \( x = 0 \):
\[ f(0) = \frac{1}{3} \]

2. At \( x = \frac{\pi}{2} \):
\[ f(\frac{\pi}{2}) = 0 \]

Thus, \( f(x) \) takes values in the interval \( [0, \frac{1}{3}] \).

### Step 9: Set the inequality for \( \lambda \)

From \( g'(x) \geq 0 \):
\[ \lambda \geq f(x) \]

Since \( f(x) \) is in \( [0, \frac{1}{3}] \), we conclude:
\[ \lambda \geq 0 \quad \text{and} \quad \lambda \leq \frac{1}{3} \]

### Conclusion

The values of \( \lambda \) such that \( g'(x) \geq 0 \) for all \( x \in [0, \frac{\pi}{2}) \) are:
\[ \lambda \in [0, \frac{1}{3}] \]